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Fundamental of Optical Engineering Lecture 8 A linearly polarized plane wave with Ē vector described by Einc ( ) eˆx eˆy is incident on an optical element under test. Describe the state of polarization of the output wave (linear, elliptrical, or circular) if the optical element is: (a) A linear polarizer oriented to transmit light polarized in the ex direction. (b) A half-wave plate with birefringence axes oriented to coincide with ex and ey. (c) A half-wave plate with birefringence axes oriented at 45º relative to ex and ey . (d) A quarter-wave plate with birefringence axes oriented to coincide with ex and ey . (e) A quarter-wave plate with birefringence axes oriented at 45º relative to ex and ey . (f) A half-wave plate with birefringence axes oriented at 25º relative to ex and ey . (g) A quarter-wave plate with birefringence axes oriented at 25º relative to ex and ey . A linearly polarized light propagating in the z-direction with polarization vector in the xdirection is incident on a birefringent crystal. What is the state of polarization of the light after passing through the crystal if: (a) the crystal is a quarter-wave plate with optic axis in the xy plane oriented at 30º relative to the yaxis? ◦ (b) the crystal is a half-wave plate with optic axis in the y-direction? ◦ (c) the crystal is a half-wave plate with optic axis in the xy plane oriented at 11º relative to the x-axis? ◦ (d) the crystal is a quarter-wave plate with optic axis in the xy plane oriented at 45º relative to the y-axis? ◦ (e) the crystal is a quarter-wave plate with optic axis in the z-direction? For a birefringent median with n0 = 1.654 and nE = 1.485 as shown in the figure. Find the length L that makes it be (a) a full wave plate (b) a half wave plate (c) a ¼ -wave plate if the wavelength is 656 nm. This makes use of electrooptic effect (applied electric fields used to change the optical properties). There are 2 kinds of electrooptic effect: linear and quadratic. The linear electrooptic effect is called “Pockels effect”. This refers to the change in the indices of the ordinary and extranordinary rays proportional to applied electric field. This effect exists only in crystals without an inversion symmetry such as LiNbO3. For a crystal with an inversion symmetry, the linear electrooptic effect can not exist, while the quadratic electrooptic effect known as “Kerr effect” is observed. This is where the induced index change is proportional to the square of applied electric field. V (pi-voltage) or half-wave voltage is the applied voltage that makes the relative phase shift be in a cube of material. In general, 2 n x n y L nx and ny = refractive index changes produced by applied voltage. V L V h It is preferable to design L >> h to have a low applied voltage V. After applying a voltage, indices are changed as nE nE nE (V ) nO nO nO (V ) V for the material is 2,700 V, L = 2 cm, and h = 0.5 mm. Find applied voltage V to have Δ = (complete extinction). An electrooptic crystal has dimensions of 2x2x3 along the x,y, and z axes with nE = 1.487 and nO = 1.536. An input wave propagating in the z-direction at λ = 0.63 μm is linearly polarized at a 45º angle relative to the x- and y- axes. A voltage V applied across the crystal in the x-direction. The voltage is increased from V = 0 until, when V = 245 V, the output polarization from the crystal is the same as that observed for V = 0. Assume that optic axis is along the yaxis. ◦ (a) What is the total phase retardation, in rad, for V = 0? ◦ (b) What is pi-voltage for the material? ◦ (c) What is the refractive index change Δnx produced by the applied voltage of 245 V, assuming Δny =0? Recall: interference eq. A1ei A2 ei 1 2 PA A12 A22 2 A1 A2 cos 1 2 Assume that n n n1 n2 1 and 2 3 1 n1 n2 n2 n3 n2 n3 i n1 n2 Er Ei e Ei n1 n2 n2 n3 4 n2 t2 round trip phase shift Er Ei A12 A23 ei ni n j amplitude of reflected wave Aij ni n j amplitude of incident wave Reflectance Er R Ei 2 A12 A23 e i 2 R A122 A232 2 A12 A23 cos We can consider R into 3 cases: ◦ n1 = n3. ◦ n1 < n2, n2>n3 , n1 ≠ n3. ◦ n1 < n2 < n3. Case 1: n1 = n3 R A122 A212 2 A12 A21 cos From a definition: Aij = Aji R 2 A122 1 cos Max in R for 2 N 1 ; N 0,1, 2,... Rmax 4 A122 Min in R for 2 N ; Rmin 0 N 0,1, 2,... n1 = 1.5, n2 = 1.6, and λ = 0.63 μm. Find t2 for Rmax and Rmin. Case 2: n1 < n2, n2>n3 , n1 ≠ n3. A12 A23 0 Max in R for 2 N 1 ; N 0,1, 2,... Rmax A12 A23 Min in R for 2 N ; 2 N 0,1, 2,... Rmin A12 A23 2 n1 = 1.5, n2 = 1.6, n3 = 1.4, λ = 0.63 μm. Find t2 for Rmax and Rmin. Case 3: n1 < n2 < n3 A12 0 A23 0 Max in R for 2 N ; N 0,1, 2,... Rmax A12 A23 Min in R for 2 N 1 ; Rmin A12 A23 2 N 0,1, 2,... 2 n1 = 1.5, n2 = 1.6, n3 = 1.7, λ = 0.63 μm. Find t2 for Rmax and Rmin.